本文整理了Java中org.opencv.core.Core.bitwise_and()
方法的一些代码示例,展示了Core.bitwise_and()
的具体用法。这些代码示例主要来源于Github
/Stackoverflow
/Maven
等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。Core.bitwise_and()
方法的具体详情如下:
包路径:org.opencv.core.Core
类名称:Core
方法名:bitwise_and
[英]Calculates the per-element bit-wise conjunction of two arrays or an array and a scalar.
The function calculates the per-element bit-wise logical conjunction for:
src1
and src2
have the same size:dst(I) = src1(I) / src2(I) if mask(I) != 0
src2
is constructed from Scalar
or has the same number of elements as src1.channels()
:dst(I) = src1(I) / src2 if mask(I) != 0
src1
is constructed from Scalar
or has the same number of elements as src2.channels()
:dst(I) = src1 / src2(I) if mask(I) != 0
In case of floating-point arrays, their machine-specific bit representations (usually IEEE754-compliant) are used for the operation. In case of multi-channel arrays, each channel is processed independently. In the second and third cases above, the scalar is first converted to the array type.
[中]计算两个数组或一个数组和一个标量的每元素逐位连接。
该函数计算以下各项的每元素逐位逻辑连接:
*src1
和src2
具有相同大小时的两个数组:
dst(I)=src1(I)/src2(I)如果掩码(I)!=0
*当src2
由Scalar
构造或具有与src1.channels()
相同数量的元素时,数组和标量:
dst(I)=src1(I)/src2如果掩码(I)!=0
*当src1
由Scalar
构造或具有与src2.channels()
相同数量的元素时,标量和数组:
dst(I)=src1/src2(I)如果掩码(I)!=0
对于浮点数组,其特定于机器的位表示(通常符合IEEE754)用于操作。对于多通道阵列,每个通道都是独立处理的。在上面的第二和第三种情况下,标量首先转换为数组类型。
代码示例来源:origin: nroduit/Weasis
public static ImageCV bitwiseAnd(Mat source, int src2Cst) {
Objects.requireNonNull(source);
ImageCV mask = new ImageCV(source.size(), source.type(), new Scalar(src2Cst));
Core.bitwise_and(source, mask, mask);
return mask;
}
代码示例来源:origin: ytai/IOIOPlotter
private void HitAndMiss(Mat src, Mat dst, Mat positive, Mat negative) {
Imgproc.erode(src, dst, positive);
Core.subtract(Mat.ones(src.size(), CvType.CV_8UC1), src, tmpMat_);
Imgproc.erode(tmpMat_, tmpMat_, negative);
Core.bitwise_and(tmpMat_, dst, dst);
}
代码示例来源:origin: kongqw/OpenCVForAndroid
public RotatedRect objectTracking(Mat mRgba) {
rgba2Hsv(mRgba);
updateHueImage();
// 计算直方图的反投影。
// Imgproc.calcBackProject(hueList, new MatOfInt(0), hist, prob, ranges, 255);
Imgproc.calcBackProject(hueList, new MatOfInt(0), hist, prob, ranges, 1.0);
// 计算两个数组的按位连接(dst = src1 & src2)计算两个数组或数组和标量的每个元素的逐位连接。
Core.bitwise_and(prob, mask, prob, new Mat());
// 追踪目标
rotatedRect = Video.CamShift(prob, trackRect, new TermCriteria(TermCriteria.EPS, 10, 1));
if (null != mOnCalcBackProjectListener) {
mOnCalcBackProjectListener.onCalcBackProject(prob);
}
// 将本次最终到的目标作为下次追踪的对象
trackRect = rotatedRect.boundingRect();
Imgproc.rectangle(prob, trackRect.tl(), trackRect.br(), new Scalar(255, 255, 0, 255), 6);
Log.i(TAG, "objectTracking: 宽度 : " + trackRect.width + " 高度 : " + trackRect.height + " 角度 : " + rotatedRect.angle);
return rotatedRect;
}
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